Two monochromatic light sources, A and B, emit the same number of photons per second. The wavelength of A is \( \lambda_A=400\,\mathrm{nm} \) and that of B is \( \lambda_B=600\,\mathrm{nm} \). The power radiated by source B is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → less than that of source A
Each photon of wavelength \( \lambda \) carries energy \( E=hc/\lambda \), so a source that emits \( N \) photons per second radiates power \( P=Nhc/\lambda \). With the same \( N \), the power is inversely proportional to the wavelength.
\( \begin{aligned} \frac{P_B}{P_A} &= \frac{\lambda_A}{\lambda_B} \\ &= \frac{400\,\mathrm{nm}}{600\,\mathrm{nm}} \\ &= \frac{2}{3} \end{aligned} \)
From the above, \( P_B=\frac{2}{3}P_A \). B emits the longer wavelength, so each of its photons carries less energy. Hence, the power of source B is less than that of source A.
NCERT (2026-27):
Particle Nature of Light: the Photon – “… (iii) All photons of light of a particular frequency ν, or wavelength λ, have the same energy E (=hν = hc/λ) and momentum p (= hν/c= h/λ), whatever the intensity of radiation may be. By increasing the intensity of light of given wavelength, there is only an increase in the number of photons per second crossing a given area, with each photon having the same energy.”